Problem: Is ${420126}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {420126}= &&{4}\cdot100000+ \\&&{2}\cdot10000+ \\&&{0}\cdot1000+ \\&&{1}\cdot100+ \\&&{2}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {420126}= &&{4}(99999+1)+ \\&&{2}(9999+1)+ \\&&{0}(999+1)+ \\&&{1}(99+1)+ \\&&{2}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {420126}= &&\gray{4\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{0\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {4}+{2}+{0}+{1}+{2}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${420126}$ is divisible by $9$ if ${ 4}+{2}+{0}+{1}+{2}+{6}$ is divisible by $9$ Add the digits of ${420126}$ $ {4}+{2}+{0}+{1}+{2}+{6} = {15} $ If ${15}$ is divisible by $9$ , then ${420126}$ must also be divisible by $9$ ${15}$ is not divisible by $9$, therefore ${420126}$ must not be divisible by $9$.